Clustering and classification

1. Read the data

In this week, we will use Boston data fom MASS to explore linear discriminant analysis (LDA) and cluster analysis. Boston data frame is about crime rate and its ralated information of Boston, USA:

• crim: perl capita crime rate by town
• zn: proportion of residential land zoned for lots over 25,000 sq.ft
• indus: proportion of non-retail business acres per town.
• chas: Charles River dummy variable (= 1 if tract bounds river; 0 otherwise).
• nox: nitrogen oxides concentration (parts per 10 million).
• rm: average number of rooms per dwelling.
• age: proportion of owner-occupied units built prior to 1940.
• dis: weighted mean of distances to five Boston employment centres.
• rad: index of accessibility to radial highways.
• tax: full-value property-tax rate per $10,000.
• ptratio: pupil-teacher ratio by town.
• black: 1000(Bk - 0.63)^2 where Bk is the proportion of blacks by town.
• lstat: lower status of the population (percent).
• medv: median value of owner-occupied homes in $1000s.

More detailed data description could be found here.

## Reading data to alc 
data('Boston')
dim(Boston)

## [1] 506  14

str(Boston)

## 'data.frame':    506 obs. of  14 variables:
##  $ crim   : num  0.00632 0.02731 0.02729 0.03237 0.06905 ...
##  $ zn     : num  18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
##  $ indus  : num  2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
##  $ chas   : int  0 0 0 0 0 0 0 0 0 0 ...
##  $ nox    : num  0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
##  $ rm     : num  6.58 6.42 7.18 7 7.15 ...
##  $ age    : num  65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
##  $ dis    : num  4.09 4.97 4.97 6.06 6.06 ...
##  $ rad    : int  1 2 2 3 3 3 5 5 5 5 ...
##  $ tax    : num  296 242 242 222 222 222 311 311 311 311 ...
##  $ ptratio: num  15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
##  $ black  : num  397 397 393 395 397 ...
##  $ lstat  : num  4.98 9.14 4.03 2.94 5.33 ...
##  $ medv   : num  24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...

Boston has 506 rows and 14 columns. chas and rad are integers, and other variables are float numbers.

2. Summary and graphical overview of the data

summary(Boston)

##       crim                zn             indus            chas        
##  Min.   : 0.00632   Min.   :  0.00   Min.   : 0.46   Min.   :0.00000  
##  1st Qu.: 0.08204   1st Qu.:  0.00   1st Qu.: 5.19   1st Qu.:0.00000  
##  Median : 0.25651   Median :  0.00   Median : 9.69   Median :0.00000  
##  Mean   : 3.61352   Mean   : 11.36   Mean   :11.14   Mean   :0.06917  
##  3rd Qu.: 3.67708   3rd Qu.: 12.50   3rd Qu.:18.10   3rd Qu.:0.00000  
##  Max.   :88.97620   Max.   :100.00   Max.   :27.74   Max.   :1.00000  
##       nox               rm             age              dis        
##  Min.   :0.3850   Min.   :3.561   Min.   :  2.90   Min.   : 1.130  
##  1st Qu.:0.4490   1st Qu.:5.886   1st Qu.: 45.02   1st Qu.: 2.100  
##  Median :0.5380   Median :6.208   Median : 77.50   Median : 3.207  
##  Mean   :0.5547   Mean   :6.285   Mean   : 68.57   Mean   : 3.795  
##  3rd Qu.:0.6240   3rd Qu.:6.623   3rd Qu.: 94.08   3rd Qu.: 5.188  
##  Max.   :0.8710   Max.   :8.780   Max.   :100.00   Max.   :12.127  
##       rad              tax           ptratio          black       
##  Min.   : 1.000   Min.   :187.0   Min.   :12.60   Min.   :  0.32  
##  1st Qu.: 4.000   1st Qu.:279.0   1st Qu.:17.40   1st Qu.:375.38  
##  Median : 5.000   Median :330.0   Median :19.05   Median :391.44  
##  Mean   : 9.549   Mean   :408.2   Mean   :18.46   Mean   :356.67  
##  3rd Qu.:24.000   3rd Qu.:666.0   3rd Qu.:20.20   3rd Qu.:396.23  
##  Max.   :24.000   Max.   :711.0   Max.   :22.00   Max.   :396.90  
##      lstat            medv      
##  Min.   : 1.73   Min.   : 5.00  
##  1st Qu.: 6.95   1st Qu.:17.02  
##  Median :11.36   Median :21.20  
##  Mean   :12.65   Mean   :22.53  
##  3rd Qu.:16.95   3rd Qu.:25.00  
##  Max.   :37.97   Max.   :50.00

From the summary table, we can find the minimun, maximum, mean, median and quantiles for each variable. Also, the observations are in difference scales. We will need to standarlize them by corresponding means and standard deviations(stds).

ggpairs(Boston, mapping = aes(col='tomato',alpha=0.05), 
        lower = list(combo =wrap("facethist", bins = 30)),
        upper = list(continuous = wrap("cor", size = 2.5)))

• The diaginal of the above figure shows the distribtion of each variable. Average number of rooms (rm) looks like normally distributed, but it is not the case for the remaining observatiopns.
• We would like to investigate the variables which can be used for crime rate prediction. From the correlation matrix in the above figure, rad, tax and lstat are the highest three variables which positively associated with crim. In contrast, medv and dis are the top two negetive associated variables with the target.
• In addition, some correlations between variables are worth to mention here, e.g. tax vs rad (0.91), idus vs nox (0.76), age vs nox (0.73), zn vs dis (0.66), rm vs medv (0.69), medv vs lstat (-0.73), dis vs age (-0.74), dis vs nox (-0.76) and lstat vs rm (-0.61).

3. Standarlization of the dataset

## standarlize Boston
boston_scaled <- scale (Boston)
## the above data is matrix, transform it to data.frame
boston_scaled <- as.data.frame(boston_scaled)
summary (boston_scaled)

##       crim                 zn               indus        
##  Min.   :-0.419367   Min.   :-0.48724   Min.   :-1.5563  
##  1st Qu.:-0.410563   1st Qu.:-0.48724   1st Qu.:-0.8668  
##  Median :-0.390280   Median :-0.48724   Median :-0.2109  
##  Mean   : 0.000000   Mean   : 0.00000   Mean   : 0.0000  
##  3rd Qu.: 0.007389   3rd Qu.: 0.04872   3rd Qu.: 1.0150  
##  Max.   : 9.924110   Max.   : 3.80047   Max.   : 2.4202  
##       chas              nox                rm               age         
##  Min.   :-0.2723   Min.   :-1.4644   Min.   :-3.8764   Min.   :-2.3331  
##  1st Qu.:-0.2723   1st Qu.:-0.9121   1st Qu.:-0.5681   1st Qu.:-0.8366  
##  Median :-0.2723   Median :-0.1441   Median :-0.1084   Median : 0.3171  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.:-0.2723   3rd Qu.: 0.5981   3rd Qu.: 0.4823   3rd Qu.: 0.9059  
##  Max.   : 3.6648   Max.   : 2.7296   Max.   : 3.5515   Max.   : 1.1164  
##       dis               rad               tax             ptratio       
##  Min.   :-1.2658   Min.   :-0.9819   Min.   :-1.3127   Min.   :-2.7047  
##  1st Qu.:-0.8049   1st Qu.:-0.6373   1st Qu.:-0.7668   1st Qu.:-0.4876  
##  Median :-0.2790   Median :-0.5225   Median :-0.4642   Median : 0.2746  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.6617   3rd Qu.: 1.6596   3rd Qu.: 1.5294   3rd Qu.: 0.8058  
##  Max.   : 3.9566   Max.   : 1.6596   Max.   : 1.7964   Max.   : 1.6372  
##      black             lstat              medv        
##  Min.   :-3.9033   Min.   :-1.5296   Min.   :-1.9063  
##  1st Qu.: 0.2049   1st Qu.:-0.7986   1st Qu.:-0.5989  
##  Median : 0.3808   Median :-0.1811   Median :-0.1449  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.4332   3rd Qu.: 0.6024   3rd Qu.: 0.2683  
##  Max.   : 0.4406   Max.   : 3.5453   Max.   : 2.9865

The data is standarlized by mean and std of each variable. As we can see, the means of the new data are zero for all of them and also are in the same scale now.

## create a quantile vector of crim from the scaled data
bins <- quantile(boston_scaled$crim)

## create a categorical variable 'crime'
crime <- cut(boston_scaled$crim, breaks = bins, label = c("low", "med_low", "med_high", "high"), include.lowest = TRUE)

## drop old crim from the dataset
boston_scaled <- dplyr::select(boston_scaled, -crim)

## add the new categorical value to scaled data
boston_scaled <- data.frame(boston_scaled, crime)

## how many rows in the scaled data
n <- nrow(boston_scaled)

## set seed to repeat the random sampling
set.seed(1111)

## randomly sample n indices between (0,n]
ind <- sample(n,  size = n)

## use the top 80% of ramndomly sampled indices as training data
train <- boston_scaled[head(ind,n=0.8*n),]
dim (train)

## [1] 404  14

## use the tail 20% of ramndomly sampled indices as test data
test <- boston_scaled[tail(ind,n=0.2*n),]
dim (test)

## [1] 102  14

4. LDA for the categorical target

# linear discriminant analysis
lda.fit <- lda(crime ~ ., data = train)

# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "pink", tex = 1, choices = c(1,2)){
  heads <- coef(x)
  arrows(x0 = 0, y0 = 0, 
         x1 = myscale * heads[,choices[1]], 
         y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
  text(myscale * heads[,choices], labels = row.names(heads), 
       cex = tex, col=color, pos=3)
}

# target classes as numeric
classes <- as.numeric(train$crime)

# plot the lda results
plot.new()
plot(lda.fit, dimen = 2,col=classes, pch=classes)
lda.arrows(lda.fit, myscale = 2)

5. Predict the target variable in test set

## save the correct classes from test data
correct_classes <- test$crime

## remove the crime variable from test data
test <- dplyr::select(test, -crime)

## predict classes with test data
lda.pred <- predict(lda.fit, newdata = test)

# cross tabulate the results
table(correct = correct_classes, predicted = lda.pred$class)

##           predicted
## correct    low med_low med_high high
##   low       18      10        0    0
##   med_low    7      14        3    0
##   med_high   1      14        9    2
##   high       0       0        0   24

From the above prediction result, we can conclude that our model is able to predict the high crime category with 100% accuracy, followed by med_low and med_high, with 69% (23/(23+3+7)) and 68% (15/(15+4+3)) accuracy. For the low crime rate, the model could make around 50% correct prediction [16/(16+15)]. Overall, the successful rate of our model is (15+23+15+16)/102 = 67.6%, and the error rate is 1-67.6% = 32.4%.

6. Clustering by K-means

## reload the data
data("Boston")
## rescale the data
boston_scaledNew <- scale(Boston)
boston_scaledNew <- as.data.frame(boston_scaledNew)
## calculate the distance matrix with default 'Euclidean distance' method.
boston_scaledNew_dist <- dist(boston_scaledNew)
summary(boston_scaledNew_dist)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##  0.1343  3.4625  4.8241  4.9111  6.1863 14.3970

The minimum and maximum Euclidean distance in boston_scaledNew are 0.13 and 14.39. The mean of the Euclidean distance is 4.91 between the observations.

## set seed to repeat the randomness in K-means
set.seed(1234)

# set the a max number of cluster number for the observations
k_max <- 10

# calculate the total within cluster sum of squares (WCSS)
twcss <- sapply(1:k_max, function(k){kmeans(boston_scaledNew, k)$tot.withinss})

# visualize the results
qplot(x = 1:k_max, y = twcss, geom = 'line')

From the above figure, we choose the n_cluster=2, because it is the elbow position which holds the ralatively lower within cluster sum of squares (WCSS) and smaller cluster number.

# k-means clustering
km <-kmeans(Boston, centers = 2)

# plot the Boston dataset with clusters
ggpairs(boston_scaledNew, mapping = aes(col=as.factor(unname(km$cluster)),alpha=0.05), 
        lower = list(combo =wrap("facethist", bins = 30)),
        upper = list(continuous = wrap("cor", size = 2.5)))

## Warning in cor(x, y, method = method, use = use): the standard deviation is
## zero

## Warning in cor(x, y, method = method, use = use): the standard deviation is
## zero

## Warning in cor(x, y, method = method, use = use): the standard deviation is
## zero

## Warning in cor(x, y, method = method, use = use): the standard deviation is
## zero

## Warning in cor(x, y, method = method, use = use): the standard deviation is
## zero

## Warning in cor(x, y, method = method, use = use): the standard deviation is
## zero

## Warning in cor(x, y, method = method, use = use): the standard deviation is
## zero

## Warning in cor(x, y, method = method, use = use): the standard deviation is
## zero

## Warning in cor(x, y, method = method, use = use): the standard deviation is
## zero

## Warning in cor(x, y, method = method, use = use): the standard deviation is
## zero

## Warning in cor(x, y, method = method, use = use): the standard deviation is
## zero

## Warning in cor(x, y, method = method, use = use): the standard deviation is
## zero

## Warning in cor(x, y, method = method, use = use): the standard deviation is
## zero